3.164 \(\int \csc ^4(e+f x) (a+b \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=82 \[ -\frac{\left (2 a^2+3 b^2\right ) \cot (e+f x)}{3 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac{a b \tanh ^{-1}(\cos (e+f x))}{f}-\frac{a b \cot (e+f x) \csc (e+f x)}{f} \]

[Out]

-((a*b*ArcTanh[Cos[e + f*x]])/f) - ((2*a^2 + 3*b^2)*Cot[e + f*x])/(3*f) - (a*b*Cot[e + f*x]*Csc[e + f*x])/f -
(a^2*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f)

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Rubi [A]  time = 0.0872102, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2789, 3768, 3770, 3012, 3767, 8} \[ -\frac{\left (2 a^2+3 b^2\right ) \cot (e+f x)}{3 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac{a b \tanh ^{-1}(\cos (e+f x))}{f}-\frac{a b \cot (e+f x) \csc (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + b*Sin[e + f*x])^2,x]

[Out]

-((a*b*ArcTanh[Cos[e + f*x]])/f) - ((2*a^2 + 3*b^2)*Cot[e + f*x])/(3*f) - (a*b*Cot[e + f*x]*Csc[e + f*x])/f -
(a^2*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f)

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \csc ^4(e+f x) (a+b \sin (e+f x))^2 \, dx &=(2 a b) \int \csc ^3(e+f x) \, dx+\int \csc ^4(e+f x) \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{a b \cot (e+f x) \csc (e+f x)}{f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x)}{3 f}+(a b) \int \csc (e+f x) \, dx+\frac{1}{3} \left (2 a^2+3 b^2\right ) \int \csc ^2(e+f x) \, dx\\ &=-\frac{a b \tanh ^{-1}(\cos (e+f x))}{f}-\frac{a b \cot (e+f x) \csc (e+f x)}{f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac{\left (2 a^2+3 b^2\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (e+f x))}{3 f}\\ &=-\frac{a b \tanh ^{-1}(\cos (e+f x))}{f}-\frac{\left (2 a^2+3 b^2\right ) \cot (e+f x)}{3 f}-\frac{a b \cot (e+f x) \csc (e+f x)}{f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x)}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.0392186, size = 132, normalized size = 1.61 \[ -\frac{2 a^2 \cot (e+f x)}{3 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac{a b \csc ^2\left (\frac{1}{2} (e+f x)\right )}{4 f}+\frac{a b \sec ^2\left (\frac{1}{2} (e+f x)\right )}{4 f}+\frac{a b \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}-\frac{a b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f}-\frac{b^2 \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sin[e + f*x])^2,x]

[Out]

(-2*a^2*Cot[e + f*x])/(3*f) - (b^2*Cot[e + f*x])/f - (a*b*Csc[(e + f*x)/2]^2)/(4*f) - (a^2*Cot[e + f*x]*Csc[e
+ f*x]^2)/(3*f) - (a*b*Log[Cos[(e + f*x)/2]])/f + (a*b*Log[Sin[(e + f*x)/2]])/f + (a*b*Sec[(e + f*x)/2]^2)/(4*
f)

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Maple [A]  time = 0.052, size = 93, normalized size = 1.1 \begin{align*} -{\frac{2\,{a}^{2}\cot \left ( fx+e \right ) }{3\,f}}-{\frac{{a}^{2}\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3\,f}}-{\frac{ab\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{f}}+{\frac{ab\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}-{\frac{{b}^{2}\cot \left ( fx+e \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(a+b*sin(f*x+e))^2,x)

[Out]

-2/3*a^2*cot(f*x+e)/f-1/3*a^2*cot(f*x+e)*csc(f*x+e)^2/f-a*b*cot(f*x+e)*csc(f*x+e)/f+1/f*a*b*ln(csc(f*x+e)-cot(
f*x+e))-1/f*b^2*cot(f*x+e)

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Maxima [A]  time = 1.99123, size = 120, normalized size = 1.46 \begin{align*} \frac{3 \, a b{\left (\frac{2 \, \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} - 1} - \log \left (\cos \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac{6 \, b^{2}}{\tan \left (f x + e\right )} - \frac{2 \,{\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a^{2}}{\tan \left (f x + e\right )^{3}}}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/6*(3*a*b*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(cos(f*x + e) - 1)) - 6*b^2/tan(f
*x + e) - 2*(3*tan(f*x + e)^2 + 1)*a^2/tan(f*x + e)^3)/f

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Fricas [A]  time = 1.89331, size = 387, normalized size = 4.72 \begin{align*} -\frac{2 \,{\left (2 \, a^{2} + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 6 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \,{\left (a b \cos \left (f x + e\right )^{2} - a b\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 3 \,{\left (a b \cos \left (f x + e\right )^{2} - a b\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 6 \,{\left (a^{2} + b^{2}\right )} \cos \left (f x + e\right )}{6 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/6*(2*(2*a^2 + 3*b^2)*cos(f*x + e)^3 - 6*a*b*cos(f*x + e)*sin(f*x + e) + 3*(a*b*cos(f*x + e)^2 - a*b)*log(1/
2*cos(f*x + e) + 1/2)*sin(f*x + e) - 3*(a*b*cos(f*x + e)^2 - a*b)*log(-1/2*cos(f*x + e) + 1/2)*sin(f*x + e) -
6*(a^2 + b^2)*cos(f*x + e))/((f*cos(f*x + e)^2 - f)*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4*(a+b*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.85136, size = 224, normalized size = 2.73 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 6 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 24 \, a b \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) + 9 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \frac{44 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 9 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 6 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{2}}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*f*x + 1/2*e)^3 + 6*a*b*tan(1/2*f*x + 1/2*e)^2 + 24*a*b*log(abs(tan(1/2*f*x + 1/2*e))) + 9*a^
2*tan(1/2*f*x + 1/2*e) + 12*b^2*tan(1/2*f*x + 1/2*e) - (44*a*b*tan(1/2*f*x + 1/2*e)^3 + 9*a^2*tan(1/2*f*x + 1/
2*e)^2 + 12*b^2*tan(1/2*f*x + 1/2*e)^2 + 6*a*b*tan(1/2*f*x + 1/2*e) + a^2)/tan(1/2*f*x + 1/2*e)^3)/f